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2 years ago

Evaluate p6 if p = 3

Mathematics

2 answers:

ANEK [815]2 years ago

6 0

18 cause p6 if p is 3 so multiple them together and get 18

HACTEHA [7]2 years ago

3 0

The answer is 18 if P=3, 3•6=18

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Find the periodic withdrawals PMT for the annuity given. HINT [See Example 4.] (Assume end-of-period withdrawals and compounding

Anettt [7]

It depends on if the interest rate of 6% is nominal or not. The question doesn't specify. Let's assume that is is a nominal rate, compounded monthly.
Start with 100000 = x * a angle 132 at .5% per month
This becomes 100000 = x * (1-1.005^-132)/.005
Which becomes (100000(.005))/(1 - 1.005^-132) = x
So x is the PMT, which is 1036.70.

8 0

2 years ago

Need this for my final... PLEASE. HELP

Blizzard [7]

Answer:

I will answer it but what is the question?

Step-by-step explanation:

7 0

2 years ago

Calculate the pH of a buffer solution made by mixing 300 mL of 0.2 M acetic acid, CH3COOH, and 200 mL of 0.3 M of its salt sodiu

Lesechka [4]

Answer:

Approximately $4.75$ .

Step-by-step explanation:

Remark: this approach make use of the fact that in the original solution, the concentration of $\rm CH_3COOH$ and $\rm CH_3COO^{-}$ are equal.

${\rm CH_3COOH} \rightleftharpoons {\rm CH_3COO^{-}} + {\rm H^{+}}$

Since $\rm CH_3COONa$ is a salt soluble in water. Once in water, it would readily ionize to give $\rm CH_3COO^{-}$ and $\rm Na^{+}$ ions.

Assume that the $\rm CH_3COOH$ and $\rm CH_3COO^{-}$ ions in this solution did not disintegrate at all. The solution would contain:

$0.3\; \rm L \times 0.2\; \rm mol \cdot L^{-1} = 0.06\; \rm mol$ of $\rm CH_3COOH$ , and

$0.06\; \rm mol$ of $\rm CH_3COO^{-}$ from $0.2\; \rm L \times 0.3\; \rm mol \cdot L^{-1} = 0.06\; \rm mol$ of $\rm CH_3COONa$ .

Accordingly, the concentration of $\rm CH_3COOH$ and $\rm CH_3COO^{-}$ would be:

$\begin{aligned} & c({\rm CH_3COOH}) \\ &= \frac{n({\rm CH_3COOH})}{V} \\ &= \frac{0.06\; \rm mol}{0.5\; \rm L} = 0.12\; \rm mol \cdot L^{-1} \end{aligned}$ .

$\begin{aligned} & c({\rm CH_3COO^{-}}) \\ &= \frac{n({\rm CH_3COO^{-}})}{V} \\ &= \frac{0.06\; \rm mol}{0.5\; \rm L} = 0.12\; \rm mol \cdot L^{-1} \end{aligned}$ .

In other words, in this buffer solution, the initial concentration of the weak acid $\rm CH_3COOH$ is the same as that of its conjugate base, $\rm CH_3COO^{-}$ .

Hence, once in equilibrium, the $\rm pH$ of this buffer solution would be the same as the ${\rm pK}_{a}$ of $\rm CH_3COOH$ .

Calculate the ${\rm pK}_{a}$ of $\rm CH_3COOH$ from its ${\rm K}_{a}$ :

$\begin{aligned} & {\rm pH}(\text{solution}) \\ &= {\rm pK}_{a} \\ &= -\log_{10}({\rm K}_{a}) \\ &= -\log_{10} (1.76 \times 10^{-5}) \\ &\approx 4.75\end{aligned}$ .

7 0

2 years ago

A grocery store manager uses 1/2 crate of apples for every 3/4 crate if oranges in a fruit display. How many crates of oranges w

ludmilkaskok [199]

Answer:

15/4

Step-by-step explanation:

This is a proportion question

It says 1/2 crates of apple are used for every 3/4 crates of oranges

For 5/2 crates of apple, x crates of oranges will be needed.

To get x, we simply multiply 1/2 by x and equate it to 3/4 multiplied by 5/2

This means: 0.5x = 0.75 × 2.5

x = ( 0.75 × 2.5 ) ÷ 0.5 = 3.75

Hence for 5/2 crates for apple, she will need 15/4 crates of oranges.

8 0

2 years ago

(-0.25x -8)-(5.5x + 9.4) the simplified expression

Alja [10]

Answer:

=−5.75x−17.4

Step-by-step explanation:

=−0.25x−8+−1(5.5x+9.4)

=−0.25x+−8+−1(5.5x)+(−1)(9.4)

=−0.25x+−8+−5.5x+−9.4

another way >>>><<<<

=−0.25x+−8+−5.5x+−9.4

=(−0.25x+−5.5x)+(−8+−9.4)

=−5.75x+−17.4

Answer:

=−5.75x−17.4

3 0

2 years ago